Question

The value of the sum $\sum _{n=1}^{13}\left(i^n+i^{n+1}\right)$ where, $i=\sqrt{-1}$ equals

• A. $i$
• B. $i-1$
• C. $-i$
• D. $0$

Answer 1

The correct option is B

$i-1$

The explanation for the correct option:

The given expression: $\sum _{n=1}^{13}\left(i^n+i^{n+1}\right)$.

It is known that, ${\mathit{i}}^{\mathbf{4}\mathbf{n}\mathbf{+}\mathbf{1}}\mathbf{=}\mathit{i}\mathbf{;}\mathbf{}{\mathit{i}}^{\mathbf{4}\mathbf{n}\mathbf{+}\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{;}\mathbf{}{\mathit{i}}^{\mathbf{4}\mathbf{n}\mathbf{+}\mathbf{3}}\mathbf{=}\mathbf{-}\mathit{i}\mathbf{;}\mathbf{}{\mathit{i}}^{\mathbf{4}\mathbf{n}}\mathbf{=}\mathbf{1}$.

Thus, $\mathit{i}\mathbf{+}{\mathit{i}}^{\mathbf{2}}\mathbf{+}{\mathit{i}}^{\mathbf{3}}\mathbf{+}{\mathit{i}}^{\mathbf{4}}\mathbf{=}\mathit{i}\mathbf{-}\mathbf{1}\mathbf{-}\mathit{i}\mathbf{+}\mathbf{1}\mathbf{=}\mathbf{0}$.

$$\begin{gathered} \sum _{n=1}^{13}\left(i^n+i^{n+1}\right)=\sum _{n=1}^{13}{i}^n\left(1+i\right) \\ \Rightarrow \sum _{n=1}^{13}\left(i^n+i^{n+1}\right)=\left(1+i\right)\sum _{n=1}^{13}{i}^n \\ \Rightarrow \sum _{n=1}^{13}\left(i^n+i^{n+1}\right)=\left(1+i\right)\left(i+i^2+i^3+......+i^{13}\right) \\ \Rightarrow \sum _{n=1}^{13}\left(i^n+i^{n+1}\right)=\left(1+i\right)\left[\left(i+i^2+i^3+i^4\right)+\left(i+i^2+i^3+i^4\right)+\left(i+i^2+i^3+i^4\right)+i\right] \\ \Rightarrow \sum _{n=1}^{13}\left(i^n+i^{n+1}\right)=\left(1+i\right)\left[0+0+0+i\right] \\ \Rightarrow \sum _{n=1}^{13}\left(i^n+i^{n+1}\right)=\left(i+i^2\right) \\ \Rightarrow \sum _{n=1}^{13}\left(i^n+i^{n+1}\right)=i-1 \end{gathered}$$

Therefore, the value of the sum $\sum _{n=1}^{13}\left(i^n+i^{n+1}\right)$ equals $i-1$.

Hence, the correct option is (B).